Unformatted text preview: 5.5.1 = − + 8 because
(d) −A + A2 = 0
2
Deﬁnition + A2 is possible = 15/22 a constant can move past1 a limit and/or a summation sign.
y 248 is 11. Theorem 5.5.4(a) depends on the fact that a constant can move past an integral sign, which by
2
y
11. Theorem 11. 5.5.1 depends onbecause athat a the can move past a past can integraly sign, which by sign, which by
5.5.4(a)
Theorem 5.5.4(a) fact constant fact can move limityand/or a summation sign.
y
Deﬁnition y y is possible thedepends on constantthat a constant an move past an integral
y
Deﬁnition 5.5.1Deﬁnition 5.5.1 is possible because a constant a limit and/or a summation sign.
is possible because a constant can move past can move past limit and/or a summation sign.
A2
12. If f (x) ≥ 0 for all x in [a, b] then we know that positivity (or nonnegativity) is preserved under
x
A2
–5
x
– A2
limits ) ≥ 0 for all2 x inalsob(by Deﬁnition 5.5.1) for integrals. nonnegativity) is preserved under
and sums, hence [a, ] then we know that positivity – 5
A2 55 x x
12. If f (x
(or
A
– 5 A1
5
A2x inknow] that positivity thatnonnegativity) nonnegativity) is preserved under
12. If f (x) ≥and sums,(x)A≥[a,also (by we [x b then we know (or A
0 1for all hence x then
–
for
a,
(or
A5
limits 12. If f x in2–1 0 b] x all Deﬁnition 5.5.1) for integrals. positivity 5 1 is preserved under
1
A
limits and1 –1 –1 limits 2and sums, Deﬁnition 5.5.1) for integrals. A1 for integrals.
alsox (by hence also (by Deﬁnition 5.5.1)
A sums, hence 4
4
A1 4
A
1
1
A1 =1 1 (3)(3) = 4 /2
13. (c) A
(a)
9
−
−A1 + A2 = − + 8 = (b)2. −A = (d) (1)(1 + 2) = −3/2
15/
−A1 + A2 = 0.
2
2
2
1
1
13. (a) A 1 (3)(3) = 9/1
=
2
(b) −A =1− (1)(1 + 2) = −3/2
1
2
13. (a) A = 13. 2 (a)= A/= (3)(3) = 9/2
92
(b) −A = y (b) −A = − 3/2 + 2) = −3/2
− (1)(1 + 2) = − (1)(1
y (3)(3)
2
2
2
22
y y2
21
1 A 1 A A 2 y
y A1 y
x 1 5x A1 A A1 6
6
6 A2 cx y
A2 x
c
A1
cAx
2 6 A2 cx x
1x 5
15. (a) −A1 + A2 = −5 + 8 = 15/2 5 = 10.
A = 2(5)
(b)(d) −A1 + A2 = 0 0; A1 = A2 by symmetry.
(c)
2
1
(c) −A1 + A2 =1− + 8 = 15/2
(d) −A1 + A2 = 0
1
10
2
(c) −A1 + A2(c)− −A18+ A2 /2 − + 8 = 15/2 (d) −A1 + A2(d)y −A1 + A2 = 0
+ = 15=
=
y=
5
2
2
y
5 y5 y 5
A1 –1 (c)
–1 –1 A1 A2 A
32 2
A
22
x
32
3 2 2 –1
2 A1 1
1y
A y
1 A x
x A2 A1
x
A1 + A2 = 1
14. (a) A = (1)(2) = 1
2
1
14. (a) A 1 (1)(2) = 1 1
= 3
2 2 –1 1
51
1
13
(5) + (1) =
.
2
22
2
2 1y x A
x –1 –11 1A
x
1 (d) 1
(b) A = (2)(3/2 + 1/2) = 2
2
1
(b) A 1 (2)(3/2 + 11 2) = 2
=
/ x
1 A= 1
[π (1)2 ] = π/2.
2 Exercise Set 5.5 139
0 0 (x + 2) dx. f (x) dx = 17. (a) −2 −2 Triangle of height 2 and width 2, above xaxis, so answer is 2.
2 0 (b) f (x) dx = 0 (2 − x) dx. (x + 2) dx + −2 −2 2 Two triangles of height 2 and base 2; answer is 4.
x − 2 dx = 0 6 2 6 (c) (2 − x) dx + 0 (x − 2) dx. 2 Triangle of height 2 and base 2 together with a triangle of height 4 and base 4, so 2 + 8 = 10.
−2 6 f (x) dx = (d) (x + 2) dx +
−2 −4 −4 6 2 0 (x + 2) dx + 0 (2 − x) dx + 2 (x − 2) dx. Triangle of height 2 and base 2, below axis, plus a triangle of height 2, base 2 above axis, another of height 2 and
base 2 above axis, and a triangle of height 4 and base 4, above axis. Thus f (x) = −2 + 2 + 2 + 8 = 10.
19. (a) 0.8 (b) −2.6 2 (c) −1.8 (d) −0.3 2 21. f (x)dx + 2
−1 −1 5 g (x)dx = 5 + 2(−3) = −1. 5 23. f (x)dx =
1 0
3 1 f (x)dx − 0 f (x)dx = 1 − (−2) = 3. 3 25. 4
−1 dx − 5 −1 1 xdx = 4 · 4 − 5(−1/2 + (3 · 3)/2) = −4. 1 27. xdx + 2
0 0 1 − x2 dx = 1/2 + 2(π/4) = (1 + π )/2. 29. False; e.g. f (x) = 1 if x > 0, f (x) = 0 otherwise, then f is integrable on [−1, 1] but not continuous.
31. False; e.g. f (x) = x on [−2, +1].
33. (a) √ x > 0, 1 − x < 0 on [2, 3] so the integral is negative. (b) 3 − cos x > 0 for all x and x2 ≥ 0 for all x and x2 > 0 for all x > 0 so the integral is positive.
35. If f is continuous on [a, b] then f is integrable on [a, b], and, considering Deﬁnition 5.5.1, for every partition and
n choice of f (x∗ ) we have
k=1 n m∆xk ≤ k=1 n f (x∗ )∆xk ≤
k k=1 n M ∆xk . This is equivalent to m(b − a) ≤ k=1 f (x∗ )∆xk ≤
k M (b − a), and, taking the limit over max ∆xk → 0 we obtain the result.
10 25 − (x − 5)2 dx = π (5)2 /2 = 25π/2. 37.
0
1 39. (3x + 1)dx = 5/2.
0 41. (a) The graph of the integrand is the horizontal line y = C . At ﬁrst, assume that C > 0. Then the region is a
b rectangle of height C whose base extends from x = a to x = b. Thus
a C dx = (area of rectangle) = C (b − a). If C ≤ 0 then the rectangle lies below the axis and its integral is the negative area, i.e. −C (b − a) = C (b − a). 140 Chapter 5 ber 10, 2008 16:01 ”ISM ET chapter 5” Sheet number 28 Page number 252 black n n ember 2008 16:01
16:01 ”ISM x) ET ,chapter 5” Sheet becomes Page
Sheet number
Page number∗ )∆ black lim
f ( ET C the 5”
( xk
ber 10, 10, 2008 (b) Since ”ISM=chapter Riema...
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